Question

P is the mid-point of the side CD of a parallelogram ABCD. A line through C parallel to PA intersects AB at Q and DA produced at R. Prove that DA = AR and CQ = QR.

Answer 1

Given: In a parallelogram ABCD, P is the mid-point of DC.

To prove: DA = AR and CQ = QR

Proof: ABCD is a parallelogram.

∴ BC = AD and BC || AD

Also, DC = AB and DC || AB

Since, P is the mid-point of DC.

∴ DP = PC = `1/2`DC

Now, QC || AP and PC || AQ

So, APCQ is a parallelogram

∴ AQ = PC = `1/2`DC

= `1/2`AB = BQ   [∵ DC = AB]  ...(i)

Now, in ΔAQR and ΔBQC,

AQ = BQ  ...[From equation (i)]

 ∠AQR = ∠BQC   ...[Vertically opposite angles]

And ∠ARQ = ∠BCQ  ....[Alternate interior angles]

∴ ΔAQR = ΔBQC   ...[By AAS congruence rule]

∴ AR = BC   ...[By CPCT rule]

But BC = DA

∴ AR = DA

Also, CQ = QR  ...[By CPCT rule]

Hence proved.